The value of $\sqrt{101}$ lies between which two consecutive integers ? Integers that appear in order when counting, for example 2 and 3.
Explanation: Consider the perfect squares near $101$ . [ What are perfect squares? Perfect squares are integers which can be obtained by squaring an integer. The first 13 perfect squares are: $ 1,4,9,16,25,36,49,64,81,100,121,144,169$ $100$ is the nearest perfect square less than $101$ $121$ is the nearest perfect square more than $101$ So, we know $100 < 101 < 121$ So, $\sqrt{100} < \sqrt{101} < \sqrt{121}$ So $\sqrt{101}$ is between $10$ and $11$.